Sorry, I was misleading in my last post. For some reason I thought you wanted the last two digits, not just the last one. To find the last digit you need only find 222^555 and 555^222 mod 10, which will be easier than finding them mod 100. This also means the bit about 11 is superfluous.

Yes, the last digit of 7^100 is 1 because they've shown that the remainder you get when you divide 7^100 by 10 is 1.

So you look at the powers of 2 and find the pattern is 4 digits long. So 2^1, 2^5, 2^9, 2^13, ...2^105, ... are all congruent to 2 mod 10. Essentially you can add a 4 to the exponent and not affect the result. Same with 2^2, 2^6,... In the language of congruences, if [tex]a\equiv b\ mod\ 4[/tex] then [tex]2^a\equiv 2^b\ mod\ 10[/tex]. There is a good reason why this is true, a power of 2's congruence class mod 10 is completely determined by its congruence class mod 5 (Chinese remainder theorem if you've met it) and we know [tex]2^4\equiv 1\ mod\ 5[/tex].